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1 RGEA Universidade de Vigo Working aper Series The equal award principle in problems with constraints and claims Gustavo Bergantiños and Leticia Lorenzo 3-06 Facultade de Ciencias Económicas e Empresariais, Campus As Lagoas-Marcosende, Vigo.

2 The equal award principle in problems with constraints and claims Gustavo Bergantiños Departamento de Estatística and Research Group in Economic Analysis. Universidade de Vigo. Leticia Lorenzo y Departamento de Estatística. Universidade de Vigo. Lagoas - Marcosende s/n Vigo, ontevedra, Spain. Abstract In this paper we study how to distribute a resource among di erent agents, claiming on it, when there are some constraints in the problem, according to the equal award principle. Some properties of the rule are given as well as an axiomatic characterization. Keywords: bankruptcy problems, ERT problems, equal award principle. Financial support from the Ministerio de Ciencia y Tecnologia and FEDER through grant SEJ C02-01 is gratefully acknowledged. y address: leticiap@uvigo.es; phone: ; fax:

3 1 Introduction Many authors have been studying the division of a resource between several agents. There is a wide literature about bankruptcy problems studied by O Neill (1982), and Aumann and Maschler (1985) among others. The reader is referred to Thomson (2003) and Moulin (2002) for a survey on this literature. Bergantiños and Sánchez (2002b) introduce ERT problems inspired by the roject Evaluation Review Technique, which is basically a method for analyzing the tasks involved in completing a given project, especially the time needed to complete each task, and identifying the minimum time needed to complete the total project, what is called ERT time. In ERT problems, there is a project that involves several activities, which have to be completed in a speci c order. The resource, extra time, is initially divided into so many pieces as paths, sets of consecutive activities from the beginning till the end of the project, each of which has a positive slack or greater time that can be added to the path without delaying the project. Once the durations of the activities are estimated, the coordinator of the project plans the schedule, computing the ERT time and nding which activities can be allowed more time if it is required. To distribute this extra time among the activities involved in the project, the coordinator would initially assign to each activity the maximum time possible, but usually that assignment would delay the project. Therefore, how should this extra time be allocated between the di erent activities with slacks? Bergantiños and Sánchez (2002a) introduce a general class of problems, which involves both bankruptcy and ERT problems, which are called problems with constraints and claims. They study how the well-known concept of proportionality can be de ned in this framework by presenting two rules, both of them based on this principle: the proportional rule (weak areto optimal) and the extended proportional rule (areto optimal). In addition, they give a characterization of the second one. The object of this paper is to study how the principle of equal award works in this kind of problems. We study two single-valued rules based on this principle: the constrained equal award rule, which satis es weak areto optimality, and the extended constrained equal award rule, which satis es areto optimality. In order to allocate the resource among the agents taking into account the claims and the new constraints, we need to nd a rule. Alternative rules represent di erent ways of applying di erent criteria to the resolution of the problem thus, which one should we choose? One way 2

4 to solve this dilemma is to identify the desirable properties that each of this rules satisfy, so that choosing a rule means choosing a set of these properties. Thus, we introduce desirable properties and check which of them are satis ed by the proposed rules. Most of these properties are inspired by those ones discussed in the literature of bankruptcy problems. However, the relations between properties in this general framework are completely di erent. For instance, properties that were compatible in bankruptcy problems, like areto optimality and composition up, become incompatible in problems with constraints and claims. We characterize the extended constrained equal award rule as the only rule satisfying areto optimality, equal treatment of equals, consistency, invariance under claims truncation, a lower bound requirement over subsets and limited composition up. The paper is organized as follows. Section 2 introduces the problems with constraints and claims and Section 3 the di erent properties of the rules. Finally, Section 4 studies the rules based on the equal award principle and characterizes the extended constrained equal award rule. 2 roblems with constraints and claims First we introduce some notation: Let N = f1; 2; :::g be the set of all possible agents. Given a nite set N N, let n be the cardinality of N. Given x = (x i ) i2n, y = (y i ) i2n 2 R N, x y means x i y i for all i 2 N; x > y means x i > y i for all i 2 N and x + y = (x i + y i ) i2n. Given S N, x S = (x i ) i2s 2 R S and x i = x Nnfig. ps We denote by = f 1 ; : : : ; p g a collection of subsets of N that covers N i=1 i = N with cardinality p. Notice that does not need to be a partition. Given M N with M 6= ;, we denote by M the collection of subsets induced by in M, i:e: M = f\m : 2 ; \M 6= ;g. Given Y R N, let B(Y ) = fy 2 Y : if x 2 R N ; x y and x 6= y then x =2 Y g be the areto boundary of Y, and W B(Y ) = fy 2 Y : if x 2 R N ; x > y then x =2 Y g the weak areto boundary. In this new kind of problems, a resource must be divided between several agents who have claims on it, as is the case for bankruptcy or taxation problems. But now, we assume that there are several subsets of agents and each subset has constraints on its awards from the point of view that the agents belonging to it can receive at most only a part of the resource available. De nition 1 A problem with constraints and claims (brie y CC) is a 4-tuple (N; ; c; E) 3

5 where N = S 2 is the set of agents, is a collection of subsets that covers N, 0 c 2 R N is the vector that holds the claims of the agents and E = (E ) 2 2 R p, where E represents an upper bound of a part of the resource available for the subset. We will often write (c; E) instead of (N; ; c; E). We denote by G(N) the class of CC with set of agents N, and by G the class of all CC. De nition 2 Given (c; E) 2 G(N), we de ne the set of feasible allocations of (c; E) by 8 9 >< F (c; E) = >: x 2 0 x i c i ; for every i 2 N >= RN x i E ; for every 2 >; : i2 A rule is a map R that associates with each problem (c; E) 2 F (c; E) an allocation R(c; E) 2 R N and where each R i (c; E) represents the award received by agent i. We now present some examples of CC problems. We prove that bankruptcy problems and ERT problems are CC problems. Finally, we give an example of a CC problem which is neither a bankruptcy problem nor a ERT problem. A bankruptcy problem (O Neill (1982)) is a 3-tuple, (N; c; E), where N denotes the set of claimants, c is the vector of claims and E c i is the resource available. We can easily i2n associate with any bankruptcy problem a CC just by considering = fng. Moreover, when the collection of subsets is a partition ( \ 0 = ; for all 6= 0 ), solving a CC can be considered as solving many bankruptcy problems as sets in the partition. We now introduce ERT problems following Bergantiños and Sánchez (2002b). Suppose that we must carry out a project in which several activities are involved, which must be completed in some speci c order. This situation is modelled by a directed graph where the arcs are the activities and each node denotes the end or the beginning of one or more activities. Each activity i has a duration d i. A path is a set of consecutive arcs from the origin to the end. The duration of a path is the sum of the durations of the activities along this path, i.e., d() = i2 d i. The ERT time is the duration of the longest path, d( ERT ) = max d(). The slack of path is ps() = d( ERT ) d() and the slack of activity i is as(i) = min ps(). A path is i2 critical if its slack is 0. We want to nish the project as soon as possible, i.e., in the ERT time. However, it is possible to assign extra time (at most ps() in path ) to some activities (at most as(i) to activity i) without delaying the project. The problem is how to do that. Formally: A ERT problem is a 3-tuple (G; (ps()) 2 ; (as(i)) i2n ) where G is the graph describing the relations between the activities, N denotes the set of activities, the set of paths, ps() is 4

6 the slack of path, and as(i) is the slack of activity i. Let us show an example to clarify the model. Example 1. B (6) A (3) C (5) D (10) This project comprises four activities (A, B, C; and D) where activity A must be completed before B and C can begin. However, there is no relation between activity D and the other activities. With each activity, or arc, is associated a number, which denotes its duration. The project has three paths: 1 = fa; Bg, 2 = fa; Cg, and 3 = fdg : Thus, ps ( 1 ) = 10 9 = 1, ps ( 2 ) = 10 8 = 2, ps ( 3 ) = = 0; as (A) = min f1; 2g = 1; as (B) = 1; as (C) = 2; and as (D) = 0. Given a ERT problem (G; (ps()) 2 ; (as(i)) i2n ), we can associate a CC (N; ; c; E), where N is the set of activities, is the set of paths, c i = as(i) for all i 2 N, and E = ps() for all 2. Note that c i is obtained from E by c i = min i2 E. Thus, ERT problems are also a subset of CC. The CC problem associated with the ERT problem of Example 1 is as follows: N = fa; B; C; Dg ; = f 1 ; 2 ; 3 g ; c = (1; 1; 2; 0) ; and E = (1; 2; 0) : There are CC problems which are neither a bankruptcy problem, nor a ERT problem. Let us show an example. Example 2. The central government of a country should divide a budget 100 among eight local governments of eight di erent regions. (8; 12; 20; 25; 30; 25; 15; 25) : The claims of the local governments are In order to divide this budget, the central government wants to take into account two characteristics of the regions: the political party governing the region and the richness of the region. Let us explain both issues more carefully. olitical parties. There are three di erent political parties 1 ; 2, and 3 in this country. Assume that party 1 is in the central government and in local governments g 1 ; g 2 ; g 3 ; 5

7 and g 4 : arty 2 is in local governments g 5 and g 6 : arty 3 is in local governments g 7 and g 8 : The central government want to avoid discrimination among the local governments in terms of the political party governing it. Then, the central government wants that for every party i, i = 1; 2; 3; the average received by the local governments of this party should not be larger than 18. Richness of the region. Assume that regions 5 and 7 are rich whereas the rest are poor. In order to help poor regions to improve their status, the central government wants to avoid that rich regions get more than 20. For each i = 1; :::; 8 let x i denote the amount received by the local government of region i: The 8 central government wants to select x = (x i ) 8 i=1 satisfying the following conditions: i:e: the budget to be divided is at most 100. i=1 x i 100; 4 x i 72; x 5 + x 6 36; and x 7 + x 8 36; i:e the average received by the local governments of each party should not be larger than 18. x 5 + x 7 20; i:e: rich regions can not receive more than 20. Thus, this problem can be model as a CC problem where N = fig 8: i=1 ; = f i g 5 i=1 with 1 = N; 2 = f1; 2; 3; 4g ; 3 = f5; 6g ; 4 = f7; 8g ; 5 = f5; 7g ; c = (8; 12; 20; 25; 30; 25; 15; 25) and E = (100; 72; 36; 36; 20) : i=1 3 roperties of the rules We devote this section to de ne some desirable properties of the rules. Most of them are wellknown properties of the rules associated with bankruptcy problems, but adapted to this general framework. We rst give an explanation of the property and then state it formally. Consider an allocation rule R. There are two well-known properties about optimality: areto optimality (O): for all (c; E) 2 G(N) we have that the award vector R(c; E) 2 B(F (c; E)), and weak areto optimality (WO): for all (c; E) 2 G(N) we have that R(c; E) 2 W B(F (c; E)). 6

8 It is logical to require that agents with equal claims, who must satisfy the same constraints, receive equal award. This property is called equal treatment of equals (ETE): for all (c; E) 2 G(N), given i; j 2 N such that c i = c j and i 2, j 2, then R i (c; E) = R j (c; E). A CC (c; E) can be interpreted as solving many subproblems as subsets in, and it would be desirable for an agent not to receive less than the minimum amount he is awarded in every subproblem in which he is involved. This is stated by the property lower bound requirement over subsets (LS): for all (c; E) 2 G(N) and every i 2 N, we have that R i (c; E) min i22 R i (c ; E ). The idea of composition up is that to divide the estate between agents is the same as dividing rst one part of the estate, in accordance with the initial claims of the agents and later, dividing the remainder with respect to the outstanding claims. Composition up (CU): for all (c; E) 2 G(N) and all E 0 2 R p such that 0 E 0 E we have that R(c; E) = R(c; E 0 ) + R(c R(c; E 0 ); E 00 ), where E 00 = E R i (c; E 0 ) for all 2. We will denote by limited i2 composition up (LCU) the property composition up restricted to problems with p = 1, i:e: = fng. The next property states that if part of a claim is above the minimum amount to divide, according with E, this part should be ignored and hence, the rule should not consider it. Invariance under claims truncation (ICT): for all (c; E) 2 G(N), R(c; E) = R(c E ; E) where n o c E i = min c i ; min E. i22 Consistency says that if some agents leave with their awards and the CC is reevaluated from the point of view of the remaining agents, the rule should allocate to these agents the same awards as those obtained in the initial problem. Consistency (CONS): for all (N; ; c; E) 2 G(N), given M N, M 6= ;; and (M; M ; c M ; EM ) 2 G(M) where E \M = E R i (N; ; c; E) i2\(nnm) for each \ M 2 M ; then, R i (M; M ; c M ; E M ) = R i(n; ; c; E) for all i 2 M. Notice that all the properties de ned above are well-known in the literature of bankruptcy except lower bound requirement over subsets (LS). Remark 1 Although the de nitions of most of these properties in CC are straightforward generalizations of properties of bankruptcy problems, the relations among these properties change completely in CC: For instance, Bergantiños and Sánchez (2002a) proved that areto optimality is incompatible with composition up, i:e: there is no rule satisfying both properties. 7

9 4 The equal award principle in CC In this section, we present two single-valued rules, both of them based on the equal award principle. This principle assigns equal amounts to all agents subject to the constraints of the problem. The constrained equal award rule for CC is de ned in a very similar way as it was in bankruptcy problems. The extended constrained equal award rule applies the equal award principle repeatedly until no payo can be increased. Both rules coincide with the constrained equal award rule when restricted to bankruptcy problems. The constrained equal award rule (brie y CEA) awards equally all agents such that no agent receives more than his claim: CEA i (c; E) = minf; c i g where = max n 0 : minf 0 ; c i g o 2 F (c; E). It can be easily proved that when c =2 F (c; E) i2n we have that = min 2 f g, where is the maximum feasible award in the subproblem (c ; E ). This rule was de ned in Sánchez (1999), who proved that CEA satis es WO and there always exists a part of the resource available which is completely allocated between its claimants, i.e. there exists 0 2 such that i2 0 CEA i (c; E) = E 0. CEA also satis es ETE and CU. Unfortunately, it fails ICT, CONS and O. So we will extend this rule such that it satis es O just by applying it repeatedly until no agent can improve his award. Before we de ne the extended rule formally, we apply this process to an example. Consider (N; ; c; E) 2 G(N) where N = f1; 2; 3g, = f 1 ; 2 g with 1 = f1; 2g and 2 = f2; 3g, c = (1; 5; 2); and E = (2; 4). The CEA awards (1; 1; 1). Note that: Agents 1 and 2 cannot be awarded more, due to the fact that for every " > 0, (1+"; 1; 1) =2 F (c; E) and (1; 1 + "; 1) =2 F (c; E), because in both cases x 1 + x 2 = 2 + " > E 1 = 2. However, agent 3 can improve his award, i:e: there exists " > 0 such that (1; 1; 1 + ") 2 F (c; E). Consider the following problem (N 2 ; 2 ; c 2 ; E 2 ), obtained when agents 1 and 2 leave the problem with their awards because their payo cannot be increased. Thus, N 2 = f3g, 2 = N 2 = ff3gg, c 2 = c 2 3 CEA 3 (c; E) = 2 1 = 1 and E 2 = E f3g = E f2;3g CEA i (c; E) = = 2. Finally, we compute CEA(c 2 ; E 2 ) = c 2 3 = 1. i=2;3 8

10 At the end, agent 1 obtains 1, agent 2 obtains 1, agent 3 obtains = 2; and (1; 1; 2) is a areto optimal feasible allocation in the problem (c; E) : We now de ne the process applied above formally: (N 1 ; 1 ; c 1 ; E 1 ) = (N; ; c; E). For any agent i 2 N we compute CEA i (c 1 ; E 1 ) = minf 1 ; c 1 i g. Assume that we have already calculated (N s ; s ; c s ; E s ) and CEA(c s ; E s ) for any s t. Let (N t+1 ; t+1 ; c t+1 ; E t+1 ) be de ned by: 8 < N t+1 = : i 2 N 9 " > 0 such that t (CEA i (c t ; E t ); CEA i (c t ; E t ) + ") 2 F (c t ; E t ) t+1 = t N t+1. 9 = ;. c t+1 = (c t i CEA i (c t ; E t )) i2n t+1. E t+1 = E t t+1 CEA t i (c t ; E t ) for any t+1 = t \ N t+1 2 t+1 with t 2 t and i2 t t \ N t+1 6= ;. For any i 2 N t+1 we compute CEA i (c t+1 ; E t+1 ) = minf t+1 ; c t+1 i g. This process ends when there exists a stage T such that N T 6= ; and N T +1 = ;. Since CEA satis es WO we have that N t+1 ends in a nite number of steps. N t for any t < T. Hence, the process described above For every agent i 2 N there exists a stage T i such that i 2 N Ti but i =2 N Ti+1. Given (c; E) 2 G(N) and i 2 N, we de ne the extended constrained equal award rule, ECEA, for agent i as T i ECEA i (c; E) = CEA i (c t ; E t ): This rule satis es O, but it is possible that for some 2 the part of the resource available is not completely allocated, i.e. i2 ECEA i (c; E) < E. Making some computations we obtain that ECEA (c; E) = (0:5; 0:5; 1:5; 0) in Example 1 and ECEA (c; E) = (8; 12; 15; 15; 10; 15; 10; 15) in Example 2. roposition 1 Given (c; E) 2 G(N), we have that that: Ti 1. ECEA i (c; E) = min t ; c i. 9

11 2. p t+1 < p t for any t = 1; : : : ; T Given i 2 N with ECEA i (c; E) < c i, there exists 2 satisfying i 2, \ N Ti 6= ;, \ N Ti+1 = ; and ECEA j (c; E) = E. roof. j2 1. Given i 2 N, when t < T i, minf t ; c t i g = t. Otherwise, minf t ; c t i g = ct i and T i = t, which is a contradiction. Thus, the award obtained by an agent i can be expressed as the sum of the awards obtained before reaching stage T i ( t when t < T i ) and the amount he is awarded at stage T i (minf Ti ; c Ti i g), i:e: T i ECEA i (c; E) = minf t ; c i g = T i 1 t=0 where we assume 0 = 0 for the sake of convenience. ( ) T i 1 t + min Ti ; c i t = min t=0 ( Ti ) t ; c i 2. It is enough to prove that for any t = 1; : : : ; T 1 there exists a subset 2 such that \ N t 6= ; and \ N t+1 = ;. Let t 2 f1; : : : ; T 1g : By de nition of CEA, there exists 2 such that t = \ N t 2 t ; t 6=?; and i2 t CEA i (c t ; E t ) = E t. Thus, for all i 2 t ; CEA i (c t ; E t ); CEA i (c t ; E t ) + " =2 F (c t ; E t ); which means that \ N t+1 = t \ N t+1 =?: 3. Given i 2 N with ECEA i (c; E) < c i, suppose that for every 2 such that i 2, \ N Ti 6= ;; and \ N Ti+1 = ; we have that ECEA j (c; E) < E. Let " > 0 be de ned in a way that j2 t=0 " < E ECEA j (c; E) for every 2 such that i 2, \ N Ti j2 \ N Ti+1 = ;. 6= ;; and " < c i ECEA i (c; E). Due to the de nition of ", CEA i c Ti ; E Ti ; CEAi c Ti ; E Ti + " 2 F c T i ; E Ti, which is a contradiction. roposition 2 ECEA satis es O, LCU, ETE, LS, CONS and ICT. 10

12 roof. It is clear that ECEA satis es O. ECEA also satis es LCU because ECEA restricted to bankruptcy problems coincides with CEA, which satis es CU. It is easy to prove that ECEA satis es ETE. ECEA satis es LS. Given an agent i 2 N, we need to prove that ECEA i (c; E) min i22 ECEA i(c ; E ): Ti By roposition 1.1, ECEA i (c; E) = min t ; c i. If Ti If Ti t c i, LS holds trivially because ECEA i (c; E) = c i : t < c i, ECEA i (c; E) = Ti t < c i. By roposition 1.3, there exists 2 where i 2, \ N Ti 6= ;, \ N Ti+1 = ; and 8 9 E = ECEA j (c; E) = < T j = min t ; c : j ; (1) : j2 j2 It is enough to prove that CEA i (c ; E ) ECEA i (c; E) because ECEA i (c ; E ) = CEA i (c ; E ): Since Ti t < c i ; c =2 F (c ; E ): Thus, CEA j (c ; E ) = minf ; c j g for all j 2 and E = j2 minf ; c j g (2) : By de nition of ECEA; T j T i for all j 2 : Thus, and (2), Ti t : Hence, T j t Ti T i ECEA i (c; E) = t minf ; c i g = CEA i (c ; E ): t for all j 2 : By (1) ECEA satis es CONS. Given (N; ; c; E) 2 G(N) and M N with M 6= ;, we must prove that ECEA i (N; ; c; E) = ECEA i (M; M ; c M ; E M ) for every i 2 M where E \M = E j2\(nnm) ECEA j (c; E) for every \ M 2 M. 11

13 We restrict the proof to the case when c =2 F (c; E). Otherwise, it is easy to prove that c M 2 F (c M ; EM ) and CONS holds trivially. We introduce some de nitions: Given i 2 M, we de ne T M i in the usual way, i 2 M T M i but i =2 M T M i +1 : T t (M) = fi 2 M : T i = tg. L(T t (M)) = fi 2 T t (M) : ECEA i (c; E) < c i g. fk r g s r=1. We take k 0 = 0: Assume that we have de ned k l for all l r 1 and k r 1 < max ig : Thus, k r = minft > k r i2m 1 : L(T t (M)) 6= ;g if ft > k r 1 : L(T t (M)) 6= ;g 6= ;: Otherwise, k r = max ig, r = s and the process ends. i2m Notice that by de nition of k r, for any agent i 2 M such that T i 2 (k r 1 ; k r ), ECEA i (c; E) = c i. We prove that given i 2 M with T i 2 (k r 1 ; k r ] we have that T M i = r and ECEA i (c M ; E M ) = ECEA i (c; E). We prove it by induction in r. We rst prove the following claim. Claim 1 Given i 2 M with T i 2 (k r kr 1 ; k r ], ECEA i (c; E) = min t ; c i. Ti roof of Claim 1. By roposition 1.1 we have that ECEA i (c; E) = min t ; c i. We distinguish two cases: 1. T i = k r. It holds trivially. 2. T i 2 (k r 1 ; k r ). By de nition of k r, ECEA i (c; E) = c i. Since T i < k r we have that kr min t ; c i = c i. Consider r = 1: Assume that CEA j (c M ; E M ) = minf1 M ; c j g for all j 2 N: The case r = 1 is a consequence of the following claims. Claim 2 1 M = k1 t. roof of Claim 2. We prove that 1 M k1 t and 1 M k1 t : 12

14 1. 1 M k1 t : It is enough to prove that k1 min t ; c j E\M for all \ M 2 M : k1 Let j 2 \ M be such that T j k 1. By Claim 1, ECEA j (c; E) = min t ; c j. ( ) Tj Let j 2 \M be such that T j > k 1. By roposition 1.1, ECEA j (c; E) = min t ; c j k1 min t ; c j. Therefore 2. 1 M k1 min ( k1 ) t ; c j ECEA j (c; E) E \M : t : Let i 2 L(T k1 (M)) be such that ECEA i (c; E) < c i. Thus, T i = k 1 : By roposition 1.3, there exists 2 with i 2, \ N k1 6= ;, \ N k1+1 = ; and ECEA j (c; E) = E. Moreover, T j k 1 for every j 2 \ M. By Claim 1, j2 Since min ( k1 ) t ; c j = minf 1 M ; c j g E\M, 1 M k1 t. ECEA j (c; E) = E \M Claim 3 If i 2 M and T i = 1 we have that T M i = 1 and ECEA i (c M ; E M ) = ECEA i(c; E): Ti roof of Claim 3. If T i < k 1, c i = ECEA i (c; E) = min t ; c i. By Claim 2, CEA i (c M ; EM ) = min k1 t ; c i : Since T i < k 1 we have CEA i (c M ; EM ) = c i: Thus, Ti M = 1 and ECEA i (c M ; EM ) = CEA i(c M ; EM ) = c i: k1 If T i = k 1 ECEA i (c; E) = min t ; c j. By roposition 1.3, there exists 2 with i 2, \ N k1 6= ;, \ N k1+1 = ; and ECEA j (c; E) = E. Moreover, T j k 1 for every j 2 \ M: By Claim 2, o min n M 1 ; c j = min j2 ( k1 ) t ; c j = ECEA j (c; E) = E \M : 13

15 Thus, T M i = 1 and ECEA i (c M ; EM ) = CEA i(c M ; EM ) = min k1 t ; c i : O RESTO DA DEMOSTRACION DIVIDEA EN VARIAS CLAIMS CO MESMO FOR- MATO QUE NO CASO ANTERIOR. TEN EN CONTA QUE ESTO AFECTARA O NUMERO DAS CLAIMS SEGUINTES. COMROBA QUE TODAS AS REFERENCIAS AS CLAIMS QUEDAN BEN. Assume the result holds for every r l. Take r = l + 1 and the problem (M l+1 ; l+1 M M l+1 = M \ N k l+1. l+1 M = ( M ) M l+1. c l+1 k l i = c i t for all i 2 M l+1. E l+1 = E \M l+1 \M = E \M by induction hypothesis. j2\(mnm l+1 ) j2\(mnm l+1 ) n Consider CEA i (c l+1 M ; (E M )l+1 ) = min In order to prove that l+1 k M l+1 for all \ M l+1 2 l+1 M l+1 min ; cl+1 M ; (E M )l+1 ) where: ECEA i (c M ; E M ) ECEA i (c; E) l+1 M k l t l+1 k l t l+1 o ; cl+1 i. We must prove that l+1 t=k l t it is enough to prove that 8 < : k l+1 t=k l +1 t ; c l+1 j 9 = ; El+1 \M l+1 M = k l+1 t=k l +1. Taking into account the de nitions of cl+1 and (E M )l+1 above, the inequality above is equivalent to the following one 8 9 < k l+1 = t ; c j : ; E \M l+1 min which we prove subsequently. = E j2\(mnm l+1 ) j2\(nnm l+1 ) roof. Given j 2 \ M l+1, we have that ECEA j (c; E) min M ECEA i (c; E) ECEA i (c; E); ( kl+1 t. t ; c j ). By roposition 1.1, it holds when T j k l+1. When k l < T j < k l+1, by Claim 1, ECEA j (c; E) = 14

16 ( kl+1 min t ; c j ). Therefore, l+1 min 8 9 < k l+1 = t ; c j : ; ECEA j (c; E) l+1 E ECEA i (c; E) j2\(nnm l+1 ) Consider i 2 L(T kl+1 (M)) with ECEA i (c; E) < c i. By roposition 1.3, there exists 2 with i 2, \ N k l+1 6= ;, \ N kl+1+1 = ; and ECEA j (c; E) = E, which implies j2 that ECEA j (c; E) = E ECEA j (c; E). Note that, by Claim 1, l+1 j2\(nnm l+1 ) ( kl+1 for every j 2 \ M l+1, k l < T j k l+1, we have that ECEA j (c; E) = min t ; c j ). Hence l+1 min which implies that 8 9 < k l+1 = t ; c j : ; = ECEA j (c; E) = E l+1 l+1 min 8 < : k l+1 t=k l +1 t ; c l+1 j Since minf l+1 M ; cl+1 i g E l+1, l+1 \M l+1 l+1 9 = M k l+1 ; = El+1 \M l+1 : t=k l +1 j2\(nnm l+1 ) ECEA i (c; E); t and hence, l+1 k M = l+1 t=k l +1 Finally, we prove that for every ( i 2 M with k l < T i k l+1 it is satis ed that Ti M = l + 1 kl+1 and ECEA i (c M ; EM ) = min t ; c i ). It holds when T i = k l+1. When T i < k l+1, ( kl+1 ECEA i (c; E) = c i. Since T i < k l+1 we have CEA i (c M ; EM ) = min t. t ; c i ) = c i = ECEA i (c M ; EM ). ECEA satis es ICT. Given (c; E) 2 G(N) and (c E ; E), where c E i = min c i ; min E i22 for all i 2 N, we must prove that ECEA(c; E) = ECEA(c E ; E). We restrict the proof to the case when c =2 F (c; E). Otherwise, c E = c and ICT holds trivially. Firstly we prove the following claims: Claim 4 Consider (N; ; c; E); (N; ; c 0 ; E) 2 G(N) where c E i c 0 i c i for all i 2 N: We de ne Ti 0 in the problem (N; ; c0 ; E) in the usual way (i 2 N T 0 i but i =2 N T 0 i +1 ). Then, given i 2 N such that T i = 1, we have that Ti 0 = 1 and ECEA i(c; E) = ECEA i (c 0 ; E). 15

17 roof of Claim???. We know that if T i = 1, ECEA i (c; E) = minf 1 ; c i g: Let 0 be such that CEA j (c 0 ; E) = minf 0 ; c 0 jg for all j 2 N. Since ECEA satis es O; for all j 2 N; ECEA j (c; E) c E j j 2 N; minf 1 ; c j g = minf 1 ; c 0 jg. Hence, for all 2 minf 1 ; c 0 jg E ; which means that 1 0. We distinguish two cases: j2 minf 1 ; c j g = j2 c 0 j c j: Thus, for all 1. ECEA i (c; E) = c i. Thus, c i 1 : Since ECEA i (c; E) c E i, ce i = c 0 i = c i: As 1 0 we have that CEA i (c 0 ; E) = c i : Thus, ECEA i (c 0 ; E) = c i and T 0 i = ECEA i (c; E) < c i : Thus, ECEA i (c; E) = 1 : By roposition 1.3, there exists 2 such that i 2, \ N 1 6= ;, \ N 2 = ; and E = j2 ECEA j (c; E) = j2 minf 1 ; c j g = j2 minf 1 ; c 0 jg: As minf 0 ; c 0 j g E, 1 0. Hence, 1 = 0, Ti 0 = 1 and j2 ECEA i (c; E) = minf 1 ; c i g = minf 1 ; c 0 ig = minf 0 ; c 0 ig = CEA i (c 0 ; E) = ECEA i (c 0 ; E): Claim 5 Consider (N; ; c; E) 2 G(N) and (M; M ; c M ; E M ) 2 G(M) with M = N t. Then given i 2 M with T i = t we have that T M i = 1. roof of Claim???. Consider i 2 M with T i = t. We know by CONS that t ECEA i (c M ; EM ) = ECEA i(c; E) = min s ; c i. Consider CEA i (c M ; EM ) = minf1 M ; c i g. s=1 We prove that 1 M = t s=1 s. t CEA i (c M ; EM ) ECEA i(c M ; EM ) = min s ; c i. Hence 1 M t s. s=1 s=1 Since M = N t, for all \ M 2 M ( t ) min s ; c i ECEA i (c; E) s=1 = ECEA i (c M ; E M ) E \M : 16

18 Therefore 1 M t s=1 s. Now we prove that ECEA satis es ICT by induction in t. By Claim 3, we have that ECEA i (c; E) = ECEA i (c E ; E) for all i 2 N such that T i = 1. We assume ECEA i (c; E) = ECEA i (c E ; E) for all i 2 N such that T i < t. Given i 2 N with T i = t, consider M = N t and the problems (M; M ; c M ; E M ) and (M; M ; c E M ; (E M )E ) associated with (N; ; c; E) and that under the induction hypothesis (E M )E = E M. By Claim 4, we know that T M i N; ; c E ; E respectively. Notice = 1. As c E j c E j c j for all j 2 M, by Claim 3 we have that ECEA i (c M ; EM ) = ECEA i(c E M ; E M ). Since ECEA satis es CONS we have that ECEA i (c; E) = ECEA i (c M ; E M ) = ECEA i (c E M ; E M ) = ECEA i (c E ; E): Theorem 1 The extended constrained equal award rule is the only rule satisfying O, LCU, ETE, LS, CONS and ICT. roof. By roposition 2, we know that ECEA satis es O, LCU, ETE, LS, CONS and ICT. Thus, to prove uniqueness, let us suppose that there exists an allocation rule R satisfying O, LCU, ETE, LS, CONS and ICT. When c 2 F (c; E) by O, R i (c; E) = c i = ECEA i (c; E). So we study the case where c =2 F (c; E). We prove the uniqueness by induction on p. Consider p = 1, i:e: = fng. In this case, ECEA(c; E) = CEA(c; E). We know, from Dagan (1996), that the constrained equal award rule is the only e cient rule satisfying ICT, CU and ETE. As R satis es O, ICT, LCU and ETE, R(c; E) = CEA(c; E) = ECEA(c; E). Assume R(c; E) = ECEA(c; E) when p k 1. Consider p = k. By LS, R i (c; E) min i22 R i(c ; E ). By induction hypothesis, R i (c ; E ) = ECEA i (c ; E ) for all 2. Since ECEA i (c ; E ) = CEA i (c ; E ) for all 2, R i (c; E) min i22 CEA i(c ; E ) = min i22 f ; c i g: Since CEA i (c; E) = minf 1 ; c i g where 1 = min 2 f g we have that R(c; E) CEA(c; E). 17

19 When T = 1, ECEA(c; E) = CEA(c; E) R(c; E). However, both rules satisfy O and therefore, R(c; E) = ECEA(c; E). So we consider T > 1. In this case, there exists 0 2 such that CEA i (c; E) = minf 1 ; c i g = E 0 : i2 0 i2 0 Hence 0 N 1 n N 2 and CEA i (c; E) = ECEA i (c; E) for every i 2 0. As R(c; E) CEA(c; E) and both rules satisfy O, we have that CEA i (c; E) = ECEA i (c; E) = R i (c; E) for all i 2 0. Consider M = N n 0 and the following problem (M; M ; c M ; EM ). As j M j = k 1, by induction hypothesis, R i (c M ; E M ) = ECEA i (c M ; E M ) for every i 2 M. Both rules satisfy CONS, therefore, R i (c; E) = ECEA i (c; E) for every i 2 M. We have just proved that R i (c; E) = ECEA i (c; E) for every i 2 0. Thus, R(c; E) = ECEA(c; E). The following table summarizes the results about CEA and ECEA (the stars mark the properties that characterize ECEA). roperties CEA ECEA WO Yes Yes O No Yes (?) ETE Yes Yes (?) LS No Yes (?) CU Yes No LCU Yes Yes (?) ICT No Yes (?) CONS No Yes (?) We end the paper proving that all the properties used in Theorem 1 but LCU are independent. We do not know what happens with this property. Let us consider R i (c; E) = 0 for every (c; E) 2 G(N) and every i 2 N. This rule satis es ETE, CONS, LS, ICT and LCU, but it does not satisfy O. Given (c; E) 2 G(N), consider R(c; E) = E RO(c; E) (Bergantiños and Sánchez 2002a). This rule satis es all properties but ICT. Consider N = f1; 2g, = fng, c = (6; 12) and E = 6. Thus, R(c; E) = (2; 4), c E = (6; 6), and R(c E ; E) = (3; 3) 6= R(c; E). 18

20 Let us de ne a priority rule R(c; E), where agents get their award according to the order 1; 2; : : : ; n; :::: Thus, given (c; E) 2 G(N) and i 2 N R i (c; E) = max 0; min nc i ; E R j (c; E)o : i22 j2:j<i Of course it satis es O, LCU (Moulin (2000)), LS, ICT and CONS. Nevertheless, R does not satisfy ETE, just consider the following example: N = f1; 2g; = N; c = (2; 2) and E = 3. Then R(c; E) = (2; 1). Let R be de ned by R i (c; E) = min i22 CEA i(c E ; E ) + E RO i (c 0 ; E 0 ) where c 0 i = c E i min CEA i(c E ; E ) i22 E 0 = E min CEA i(c E ; E ) i22 i2 Of course R satis es O and ICT by de nition. Moreover it satis es LCU because when j j = 1, it coincides with CEA which satis es composition up. Since both CEA and ERO satisfy ETE, R also satis es ETE. Furthermore it satis es LS because it can be proved that min i22 R i(c ; E ) min i22 CEA i(c E ; E ) R i (c; E). Moreover, it does not satisfy CONS. Consider the following example. N = f1; 2; 3; 4g, 1 = f1; 2; 3; 4g, 2 = f1; 2g, E = (20; 2) and c = (2; 2; 10; 15). Thus, R(c; E) = 1; 1; 76 9 ; Given M = f3; 4g, M = fmg, EM = (18) and c M = (10; 15). Then, R(c M ; EM ) = 76 (9; 9) 6= 9 ; Given (c; E) 2 G(N), consider the following relation between agents: i j, for every 2, i 2, j 2. We denote by [k] the equivalence class of agent k 2 N. We associate with every [k] 2 N e a vector x k = x k 2 2 Rp such that: 8 < x k 0 if [k] \ = ; = : 1 if [k] We de ne the rule in two stages: S1 We divide E among the equivalence classes following the priority rule given by the lexicographic order. If x i and x k are the vectors associated with the equivalence classes [i] and [k], we say that [i] has the priority over [k] if and only if x i L x k, where L denotes the lexicographic order. We denote by R[k] L (c; E) the award received by the equivalence class [k] in this stage. 19

21 S2 We divide R[k] L (c; E) among the members of [k] following the equal award principle. Then, for any i 2 [k]: Ri L(c; E) = CEA i(c [k] ; R[k] L (c; E)). R clearly satis es O and LS. It also satis es ETE because the CEA rule satis es ETE. ICT is satis ed since both the priority rule and the CEA rule satisfy this property. Moreover it satis es LCU because when j j = 1, R coincides with the CEA rule. However, R does not verify LS. Consider the following example where N = f1; 2; 3g, = f 1 ; 2 g with 1 = f1; 2g and 2 = f2; 3g, E = (4; 4) and c = (4; 8; 4). The lexicographic order is [2], [1] and [3]. If we compute R L (c; E) = (0; 4; 0). Although for agent 1, R L 1 (c ; E ) = R1 L (c 1 ; E 1 ) = CEA 1 (c 1 ; E 1 ) = 2 > 0 = R1 L (c; E). min References Aumann, R.J., Maschler, M., Game theoretic analysis of a bankruptcy problem from the Talmud, Journal of Economic Theory Bergantiños, G., Sánchez, E., 2002a. The proportional rule for problems with constraints and claims, Mathematical Social Sciences Bergantiños, G., Sánchez, E., 2002b. NTU ERT games, Operations Research Letters 30 (2) Dagan, N., New characterizations of old bankruptcy rules, Social Choice and Welfare Moulin, H., riority rules and other asymmetric rationing methods, Econometrica 68 (3) Moulin, H., Axiomatic cost and surplus sharing. In Arrow, K.J., Sen, A.K., Suzumura, K. (Eds.) Handbook of Social Choice and Welfare, Vol 1, pp O Neill, B., A problem of rights arbitration from the Talmud, Mathematical Social Sciences Sánchez, E., Juegos cooperativos que describen modelos en los que el orden es inherente al problema (in Spanish), Universidade de Santiago de Compostela (Eds.). Thomson, W., Axiomatic and Game-Theoretic analysis of Bankruptcy and Taxation problems: a survey. Mathematical Social Sciences